Inequalities regarding the geometric dimension of groups

105 Views Asked by At

I'm studying some group cohomology, and I'm stuck on the following problem (namely, problem VIII.1.3 of Brown's Cohomology of Groups): let's define the geometric dimension of a group $G$ as $$ \text{gd}G = \inf\{n \, | \, \dim_{CW}(X), \, X \text{ a classifying space for }G \}, $$ where we take as classifying spaces the $K(G,1)$'s that are also path-connected CW-complexes.

Similarly to the case of cohomological dimension (where the cohomological dimension of $G$ is defined as the infimum of the length of projective resolutions of $\mathbb{Z}$ over $\mathbb{Z} G$), I would like to show the following:

  1. if $A \subset G$ is a group inclusion, then $$ \text{gd} A \leq \text{gd} G; $$
  2. If $1 \to G_1 \to G \to G_2 \to 1$ is a group extension, then $$ \text{gd} G \leq \text{gd} G_1 + \text{gd} G_2; $$
  3. If $G = G_1 \ast_A G_2$ where $A \subset G_1, G_2$, then $$ \text{gd} G \leq \max \{\text{gd} G_1, \text{gd} G_2, 1 + \text{gd} A\}. $$

My attempt:

For 1: this is more or less clear. Any classifying space $X$ as above admits an universal cover $\tilde{X} \to X$; by the classification theorem of connected covers of $X$, we get another cover $p_A: Y \to X$ such that ${p_{A}}_\ast(\pi_1(Y)) = A \subset G = \pi_1(X)$, and the induced map $\tilde{X} \to Y$ is also a cover.

I claim that $Y$ is a classifying space: first of all, since ${p_A}_\ast$ is injective, $\pi_1(Y) \cong A$. Moreover, $\tilde{X}$ is a CW complex (by lifting the CW-structure from $X$) such that $\dim_{CW}(\tilde{X}) = \dim_{CW}(X)$, and $\pi_{\geq 2}(\tilde{X}) = \pi_{\geq 2}(X) = 0$; by the same reasoning $Y$ is a CW-complex, $\pi_{\geq 2}(\tilde{Y}) = \pi_{\geq 2}(Y) = 0$ and we have that $\dim(Y) \leq \dim(X)$.

For 2: here I'm stuck. I know I can get a fibre sequence $K(G_1,1) \to K(G,1) \to K(G_2,1)$, and then use the Serre spectral sequence to get, for any abelian group $M$, $$ H^p(K(G_2,1);H^q(K(G_1,1);M)) \Rightarrow H^{p+q}(K(G,1);M), $$ and for $k> \text{gd}G_2 + \text{gd} G_1$, by taking $p+q =k$ and by the vanishing of the second page, I get that $H^{k}(K(G,1); M) \cong 0$. Is this enough to conclude? If yes, how?

For 3: also here I'm stuck, similarly as before. Briefly, the idea I have is: take $X_1$ and $X_2$ classifying spaces for $G_1$ and $G_2$ respectively, minimizing the dimensions. Since $A \subset G_1, G_2$ and $\text{Hom}(A, G_i) \cong [K(A,1), K(G_i,1)]_\ast$, we can find maps from $X_A$, the $K(A,1)$ minimizing the dimension for $A$, fitting in a diagram of cellular maps: $$ X_1 \leftarrow X_A \rightarrow X_2 $$

By taking the pushout and calling it $X$, by van Kampen and Mayer-Vietoris (applied on the universal cover), this should be a classifying space for $G$. I guess that at this point (at least intuitively) $\text{gd} G \leq \max\{\text{gd} G_1, \text{gd} G_2 \}$, but I can't get the dimension of $A$ coming into play here (maybe I should use some sort of Mayer-Vietoris technique here)? Also, does this make sense, even though it's really sketchy?

Any help is appreciated. Thanks!